Legendre transformation without Legendre transformation

Aug 12, 2013 • Alex Rogozhnikov

If you heard something of theoretical mechanics, you definitely
know that the most important transition in mechanics is one from Lagrangian Mechanics to Hamiltonian. And its name is
Legendre transform.

This transition is some kind of magic that to my mind is considered usually as some
secret recipe. The recipe is

And that is all the idea. But why should I take this Hamiltonian not some other function? Why
should I consider Lagrangian derivatives as new variables instead of $q_i$ ? I didn't meet the answers in
mechanics courses, though there is one simple intuitive justification. That's Pasha Gavrilenko who told me about
it.

What do we have initially? An action on some interval of time, this is the integral

$$ S = \min_{q(\cdot)} \int_{t_1}^{t_2} L(q, \dot{q}, t) dt$$ which should be minimized (locally), that
what Hamilton's principle states. (There are conditions on the endpoints which I will omit)

See? We have now $q$ and $v$ independent by the cost of
additional summand. Now we can write $\delta$ in the following form (make sure you understand it):
$$\delta[q,v] = \max_{p(\cdot)} \int_{t_1}^{t_2} p(t) (\dot{q}(t) - v(t)) dt $$

After substitution we have the problem on finding saddle point of the function:
$$ S = \min_{q(\cdot), v(\cdot)} \max_{p(\cdot)} \int_{t_1}^{t_2} p(t) (\dot{q}(t) - v(t)) + L(q, v, t) dt $$
Pay attention that all variables $p,q,v$ are independent now.
The solution we need is trajectory $q(\cdot)$, but it has corresponding trajectories $v(\cdot)$ and $p(\cdot)$ which form a saddle point together with $q(\cdot)$.